User blog:B1mb0w/Fundamental Sequences
'Fundamental Sequences' This blog will map out much of the standard definitions on Fundamental Sequences for Ordinals. It will attempt to simplify the rule-set where possible. The alternative rule-set presented here will be used in my other blog for the J Function and in particular in the Sandpit \(J_4\) blog. 'Basics (Cantor's Normal Form)' Let \(\gamma\) and \(\delta\) be two arbitrary transfinite ordinals, and \(n\) is a finite integer. Then: \((\gamma + 1)n = \gamma\) \((\gamma + \delta)n = \gamma + \deltan\) when \(\gamma >> \delta\) \(\gamma.(\delta + 1)n = \gamma.\delta + \gamman\) \(\gamma.\deltan = \gamma.(\deltan)\) when \(\gamma >> \delta\) \(\gamma^{\delta + 1}n = \gamma^{\delta}.(\gamman)\) and \(\gamma^{\delta}n = \gamma^{\deltan}\) 'Some Common Transfinite Ordinals' \(\omegan = n\) \(\epsilon_0n = \omega\uparrow\uparrow n\) \(\epsilon_1n = \epsilon_0\uparrow\uparrow n\) \(\epsilon_{j+1}n = \epsilon_j\uparrow\uparrow n\) and \(\epsilon_{\omega}n = \epsilon_{\omegan} = \epsilon_n\) 'Veblen Hierarchy' Continuing into Veblen Hierarchy and the \(\varphi\) function. Lets start with these equations which are equivalent to those in the Common Transfinite Ordinal section. \(\varphi(1)n = \omegan = n\) \(\varphi(1,0)n = \epsilon_0n = \varphi(n) = \omega\uparrow\uparrow n\) \(\varphi(1,1)n = \epsilon_1n = \varphi(1,0)\uparrow\uparrow n\) \(\varphi(1,j + 1)n = \epsilon_{j + 1}n = \varphi(1,j)\uparrow\uparrow n\) and \(\varphi(1,\omega)n = \varphi(1,\omegan) = \varphi(1,n)\) '(1) Rule-set' At this point, we can focus on the commonly used rule-sets and generalise all the above and more. Let's allow \(\alpha\) and \(\beta\) to be transfinite ordinals. Then the following rules seem to apply but they do not appear to be self-consistent. \(\varphi(\alpha,\beta)n = \varphi(\alpha,\betan)\) \(\varphi(\alpha,\beta + 1)n = \varphi(\alphan,\varphi(\alpha,\beta)+1)\) \(\varphi(\alpha,0)n = \varphi(\alphan,0)\) and \(\varphi(\alpha + 1,0)0 = 0\) \(\varphi(\alpha + 1,0)+ 1 = \varphi(\alpha,\varphi(\alpha + 1,0)n + c)\) where \(c = 0\) or \(1\) ??? and \(\varphi(\alpha + 1,\beta + 1)0 = \varphi(\alpha + 1,\beta) + 1\) but why ??? \(\varphi(\alpha + 1,\beta + 1)+ 1 = \varphi(\alpha,\varphi(\alpha + 1,\beta + 1)n)\) And, the following rules do not seem to be consistent with the general rules just given. \(\varphi(0,\beta)n = \varphi(0,\betan) = \varphi(\betan)\) \(\varphi(0,\beta + 1)n = \varphi(\beta + 1)n = \varphi(\beta)^n\) 'A Question' If we focus on this rule: \(\varphi(\alpha,\beta + 1)n = \varphi(\alphan,\varphi(\alpha,\beta)+1)\) Then why is the next rule so different: \(\varphi(\alpha + 1,\beta + 1)+ 1 = \varphi(\alpha,\varphi(\alpha + 1,\beta + 1)n)\) Lets compare them: Let \(\gamma = \alpha + 1\) \(\varphi(\gamma,\beta + 1)n = \varphi(\gamman,\varphi(\gamma,\beta)+1)\) then \(\gamman = (\alpha+1)n = \alpha\) and then we get a different result \(\varphi(\alpha + 1,\beta + 1)n = \varphi(\alpha,\varphi(\alpha+1,\beta)+1)\) 'Calculated Example' What is the fundamental sequence for \(\zeta_02\) ? \(\zeta_02 = \varphi(2,0)2 = \varphi(1,\varphi(2,0)1 + c)\) \(= \varphi(1,\varphi(1,\varphi(2,0)0 + c) + c) = \varphi(1,\varphi(1,0 + c) + c)\) then if c = 0 \(\zeta_02 = \varphi(2,0)2 = \varphi(1,\varphi(1,0)) = \epsilon_{\epsilon_0}\) of if c = 1 \(\zeta_02 = \varphi(2,0)2 = \varphi(1,\varphi(1,1) + 1) = \epsilon_{\epsilon_1+1}\) '(2) Alternative Rule-set' This alternative rule set is more succinct. Lets start with some conventions as follows: \(k^2(n,p_*) = k(n,k(n,p))\) \(k^2(n_*,p) = k(k(n,p),p)\) and \(k(a_{2},b_{3}) = k(a_1,a_2,b_1,b_2,b_3)\) then Rule 1: \(\varphi(1,0_{m+1}) = \varphi^{\omega}(1_*,0_{m})\) Rule 2: \(\varphi(\alpha+1,0_{m+1}) = \varphi^{\omega}(\alpha,0_*,0_{m})\) Rule 3a: \(\varphi(\mu_{m},\beta+1) = \varphi(\mu_{m},\beta)\uparrow\uparrow\omega\) where either \(\mu\) or m are non-zero, or otherwise: Rule 3b: \(\varphi(0,n) = \varphi(n) = \omega^n\) The rules are diagonalised as follows: \(\varphi(1,0_{m+1})n = \varphi^n(1_*,0_{m})\) \(\varphi(\alpha+1,0_{m})n = \varphi^n(\alpha_*,0_)\) \(\varphi(\alpha,0_{m+1})n = \varphi^n(1_*,0_{m})\) \(\varphi(\alpha,\beta+1)n = \varphi(\alpha,\beta)\uparrow\uparrow n\) 'Is Rule 3 correct ?' It has been pointed out in the comments that: \(\beta\uparrow\uparrow\omega = \varphi(1,\beta+1)\) This cannot be correct because of the following: \(\epsilon_0\uparrow\uparrow\omega = \epsilon_1 = \varphi(1,1)\) \(\epsilon_{\delta}\uparrow\uparrow\omega = \epsilon_{\delta + 1} = \varphi(1,\delta + 1)\) Therefore, it is the "subscript of epsilon" that can be input into the \(\varphi\) function on the right hand sand. But what happens if the "subscript of epsilon" for \(\beta = \zeta_0\) is the same as \(\beta\), i.e.: \(\zeta_0\uparrow\uparrow\omega = \epsilon_{\zeta_0}\uparrow\uparrow\omega = \varphi(1,\zeta_0 + 1)\) Then we have a different problem, because I argue \(\epsilon_{\zeta_0}\) is not well defined. Lets start with this agreed definition: \(\zeta_0 = \varphi^{\omega}(1,0_*) = \varphi^{1 + \omega}(1,0_*) = \varphi(1,\varphi^{1 + \omega}(1,0_*)) = \varphi(1,\zeta_0)\) So far so good, but what when you evaluate these expressions for say the diagonal of 2 ? \(\zeta_02 = \varphi^{\omega}(1,0_*)2 = \varphi^2(1,0_*) = \varphi(1,\varphi(1,0))\) and then \(\varphi(1,\zeta_0) = \varphi(1,\varphi^{\omega}(1,0_*))2 = \varphi(1,\varphi^2(1,0_*)) = \varphi(1,\varphi(1,\varphi(1,0)))\) This is a significant contradiction. The cause of the contradiction in my opinion is allowing constructed ordinals of the form \(\epsilon_{\zeta_0}\). We cannot guarantee that these are understood or well defined. 'Calculated Example' What is the fundamental sequence for \(\zeta_02\) using the alternative rule set ? \(\zeta_02 = \varphi(2,0)2 = \varphi_{1+1}(0)2\) \(= \varphi^{\omega}(1,0_*)2 = \varphi^2(1,0_*) = \varphi(1,\varphi(1,0)) = \epsilon_{\epsilon_0}\) This result is the same as the previous calculated example when c = 0, but the calculation is actually simpler than that used in the previous calculated example, because there was no need to include the terms: \(\varphi(2,0)1\) and \(\varphi(2,0)0\) And also no need for the additional rules defined in the original rule set, such as: \(\varphi(\alpha + 1,0)0 = 0\) I have created another blog to calculate \(f_{\zeta_0}(2)\) in detail. '\(\Gamma_0\)' Calculating \(\Gamma_0\) we get: \(\varphi(1,0,0) = \Gamma_0\) then \(\Gamma_02 = \varphi(1,0,0)2 = \varphi^{\omega}(1_*,0)2 = \varphi^2(1_*,0) = \varphi(\varphi(1,0),0)\) 'Small Veblen Ordinal (SVO)' SVO is defined as follows: \(SVO = \varphi(1,0_{\omega})\) Diagonalising SVO for n=2 produces this result: \(SVO2 = \varphi(1,0_{\omega})2 = \varphi(1,0_{2}) = \varphi(1,0,0) = \Gamma_0\) ' Appreciate any comments on this blog.' Category:Blog posts